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Conformal Scaling of the Ten-Dimensional Lorentzian Hodge Star

Statement

Let $g$ be an oriented Lorentzian metric of inertia $(9, 1, 0)$ on a smooth $10$ -manifold, let $ω$ be a smooth real function, and set $\hat{g} = e^{2 ω} g$ . For every smooth $p$ -form $β$ , $0 \leq p \leq 10$ ,

$*_{\hat{g}} β = e^{(10 - 2 p) ω} *_{g} β .$

In particular, the Hodge star on $5$ -forms is unchanged by this conformal rescaling.

Statement references

Lorentzian Hodge Star in Ten Dimensions

Existence and Five-Form Properties of the Lorentzian Hodge Star

Facts & Assumptions

Given: The data and conformal rescaling in the Statement.

[F1]

Lorentzian Hodge Star in Ten Dimensions: the star is uniquely characterized by $α \land *_{g} β = ⟨ α, β ⟩_{g} {vol}_{g}$ .

[F2]

Existence and Five-Form Properties of the Lorentzian Hodge Star: the metric volume form and Hodge star are well-defined and unique.

Proof

1.1F1F2algebra

In a positive coordinate chart, ${\hat{g}}_{μ ν} = e^{2 ω} g_{μ ν}$ , so $\det (\hat{g}) = e^{20 ω} \det (g)$ and the positive square root of its absolute value gives ${vol}_{\hat{g}} = e^{10 ω} {vol}_{g}$ .

1.2F1givenalgebra

Raising the $p$ covector indices in the pairing uses $p$ inverse metric factors, each scaled by $e^{- 2 ω}$ . Hence $⟨ α, β ⟩_{\hat{g}} = e^{- 2 p ω} ⟨ α, β ⟩_{g}$ .

2.1F1F2step 1.1step 1.2algebra∎

By [F1] and steps 1.1-1.2, for every $p$ -form $α$ we have $α \land *_{\hat{g}} β = e^{(10 - 2 p) ω} α \land *_{g} β = α \land (e^{(10 - 2 p) ω} *_{g} β)$ . The wedge pairing with all $α$ is nondegenerate by the complementary-basis argument in [F2], so the two $(10 - p)$ -forms paired with $α$ are equal. This proves the formula. For $p = 5$ its exponent is zero.

Depends on

Dependency tree · 2 levels

Sources

Sean M. Carroll, Lecture Notes on General Relativity, arXiv:gr-qc/9712019v1